[leetcode] Insert Interval
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals[1,3],[6,9], insert and merge[2,5]in as[1,5],[6,9].
Example 2:
Given[1,2],[3,5],[6,7],[8,10],[12,16], insert and merge[4,9]in as[1,2],[3,10],[12,16].
This is because the new interval[4,9]overlaps with[3,5],[6,7],[8,10].
https://oj.leetcode.com/problems/insert-interval/
思路1:建立新的结果返回,遍历原来的线段组(注意原来是按照start排序之后的),所以根据是否和新的interval重叠,加入新的结果中(结果也要求按start排序)。
public class Solution { public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { if (newInterval == null) return intervals; ArrayList<Interval> res = new ArrayList<Interval>(); for (Interval each : intervals) { if (each.end < newInterval.start) res.add(each); else if (each.start > newInterval.end) { res.add(newInterval); newInterval = each; } else { newInterval = new Interval(Math.min(each.start, newInterval.start), Math.max(each.end, newInterval.end)); } } res.add(newInterval); return res; } public static void main(String[] args) { // Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as Interval one = new Interval(1, 2); Interval two = new Interval(3, 5); Interval three = new Interval(6, 7); Interval four = new Interval(8, 10); Interval five = new Interval(12, 16); ArrayList<Interval> intervals = new ArrayList<Interval>(); intervals.add(one); intervals.add(two); intervals.add(three); intervals.add(four); intervals.add(five); System.out.println(new Solution().insert(intervals, new Interval(4, 9))); } }
第二遍记录:
注意各种情况的处理, 最后结果也要保持原有顺序。
参考:
http://www.programcreek.com/2012/12/leetcode-insert-interval/
http://www.cnblogs.com/TenosDoIt/p/3715013.html